If the point (2, -3) lies on $k x^{2} - 3 y^{2} + 2 x + y - 2 = 0$ then k is equal to

\frac{1}{7}

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Solution

The correct option is C 7
As the point lies on the given line, it should satisfy the equation of the line, if we substitute $x = 2$ and $y = - 3$ in it.
So, $k ({2)}^{2} 3 {(- 3)}^{2} + 2 (2) - 32 = 0$
$=> 4 k - 27 + 4 - 3 - 2 = 0$
$4 k = 28$
$k = 7$

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